使用 jq 重新格式化 json - 将结果收集到数组中

Question

我有一个返回 json 的查询，如下所示：

{
  "hostname": "seapr1pgdb032",
  "ip": "10.215.0.6",
  "dataCenter": "seapr1"
}
{
  "hostname": "seapr1cpndb001",
  "ip": "10.203.0.41",
  "dataCenter": "seapr1"
}
{
  "hostname": "seapr1dhcp01",
  "ip": "10.205.3.212",
  "dataCenter": "seapr1"
}

数据中心可以变化，我想将每个数据中心的所有主机收集到一个对象中，如下所示：

{
 "dataCenter": "seapr1",
 "hosts": [
    "seapr1pgdb032",
    "seapr1cpndb001",
    "seapr1dhcp01"
  ] 
}

在Reshaping JSON with jq 工作，我认为这样做可以：

{dataCenter: .dataCenter, hosts: [.hostname] } 

但我得到了三个数据中心对象，而不是我预期的单个合并对象：

{
  "dataCenter": "seapr1",
  "hosts": [
    "seapr1pgdb032"
  ]
}
{
  "dataCenter": "seapr1",
  "hosts": [
    "seapr1cpndb001"
  ]
}
{
  "dataCenter": "seapr1",
  "hosts": [
    "seapr1dhcp01"
  ]
}

Answer 1

这是一个 jq 解决方案，它避免了内置的 group_by 过滤器，原因在下面的 cmets 中解释。

# This stream-oriented variant of the built-in `group_by` emits a stream of arrays,
# each array representing a group. The main advantages over `group_by` are
# that no sorting is required, and the ordering of items within a group is preserved.
# There are no restrictions on f or the items in the stream;
# in particular, f may evaluate to any number of values,
# but if f evaluates to empty at any item in the stream, then that item will in effect be discarded.
# Example:
#  GROUPS_BY( 4,3,2,1; . % 2 ) => [4,2] [3,1]

def GROUPS_BY(stream; f): 
  reduce stream as $x ({};
    reduce [$x|f][] as $s (.;
        ($s|type) as $t
        | (if $t == "string" then $s
           else ($s|tojson) end) as $y
        | .[$t][$y] += [$x] ) ) 
  | .[][] ;

使用 -n 命令行选项，我们可以通过编写如下解决方案来避免“吞咽”输入：

GROUPS_BY(inputs; .dataCenter)
| {dataCenter: .[0].dataCenter, hosts: map(.hostname) }